From a point $A$,$h \text{ m}$ above the ground level,the angle of elevation of the top of a tower is $\alpha$ and the angle of depression of the base of the tower is $\beta$. Prove that the height of the tower is $\frac{h(\tan \alpha + \tan \beta)}{\tan \beta} \text{ m}$.

(N/A) Let $\overline{CD}$ be the tower and $A$ be the point of observation $h \text{ m}$ above the ground level.
Let $\overline{AE} \perp \overline{CD}$,where $E$ lies on $\overline{CD}$.
Then,$\angle DAE = \alpha$,$\angle EAC = \beta$,and $AB = h \text{ m}$.
Let $CD = x \text{ m}$ and $BC = y \text{ m}$.
Then $AE = BC = y \text{ m}$ and $CE = AB = h \text{ m}$.
Also,$DE = DC - CE = (x - h) \text{ m}$.
In $\Delta ABC$,$\angle B = 90^{\circ}$.
$\therefore \tan \beta = \frac{AB}{BC} = \frac{h}{y} \implies y = \frac{h}{\tan \beta} \quad \dots(1)$
In $\Delta DEA$,$\angle E = 90^{\circ}$.
$\therefore \tan \alpha = \frac{DE}{AE} = \frac{x - h}{y} \implies y = \frac{x - h}{\tan \alpha} \quad \dots(2)$
From $(1)$ and $(2)$:
$\frac{h}{\tan \beta} = \frac{x - h}{\tan \alpha}$
$h \tan \alpha = (x - h) \tan \beta$
$h \tan \alpha = x \tan \beta - h \tan \beta$
$x \tan \beta = h \tan \alpha + h \tan \beta$
$x \tan \beta = h(\tan \alpha + \tan \beta)$
$x = \frac{h(\tan \alpha + \tan \beta)}{\tan \beta}$
Thus,the height of the tower is $\frac{h(\tan \alpha + \tan \beta)}{\tan \beta} \text{ m}$.